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I'm going to assume that the room is a rectangle.
The area of a rectangle is A = lw, where l=length of the rectangle and w=width of the rectangle.
You're given that the length, l = (x+5)ft and the width, w = (x+4)ft. You're also told that the area, A = 600 sq. ft. Plug these values into the equation for the area of a rectangle and FOIL to multiply the two factors:
Now subtract 600 from both sides to get a quadratic equation that's equal to zero. That way you can factor the quadratic to find the roots/solutions of your equation. One of the solutions is the value of x that you would use to find the dimensions of the room:
Now you know that x could be -29 or 20. For dimensions, the value of x must give you a positive value for length and width. That means x can only be 20. Plugging x=20 into your equations for the length and width, you get:
Length = x + 5 = 20 + 5 = 25 ft.
Width = x + 4 = 20 + 4 = 24 ft.
The dimensions of your room are 25ft (length) by 24ft (width).
The area of a rectangle is A = lw, where l=length of the rectangle and w=width of the rectangle.
You're given that the length, l = (x+5)ft and the width, w = (x+4)ft. You're also told that the area, A = 600 sq. ft. Plug these values into the equation for the area of a rectangle and FOIL to multiply the two factors:
Now subtract 600 from both sides to get a quadratic equation that's equal to zero. That way you can factor the quadratic to find the roots/solutions of your equation. One of the solutions is the value of x that you would use to find the dimensions of the room:
Now you know that x could be -29 or 20. For dimensions, the value of x must give you a positive value for length and width. That means x can only be 20. Plugging x=20 into your equations for the length and width, you get:
Length = x + 5 = 20 + 5 = 25 ft.
Width = x + 4 = 20 + 4 = 24 ft.
The dimensions of your room are 25ft (length) by 24ft (width).
The limit as a definite integral on the interval on [2π , 4π] is
.
<h3>What is meant by definite integral?</h3>
A definite integral uses infinitesimal slivers or stripes of the region to calculate the area beneath a function. Integrals can be used to represent a region's (signed) area, the cumulative value of a function changing over time, or the amount of a substance given its density.
Definite integral, a term used in mathematics. is the region in the xy plane defined by the graph of f, the x-axis, and the lines x = a and x = b, where the area above the x-axis adds to the total and the area below the x-axis subtracts from the total.
If an antiderivative F exists for the interval [a, b], the definite integral of the function is the difference of the values at points a and b. The definite integral of any function can also be expressed as the limit of a sum.
Let the equation be
substitute the values in the above equation, we get
= on [2π, 4π],
simplifying the above equation
To learn more about definite integral refer to:
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